Let the body travel the height $h$ in time $t$. Since the body is dropped, initial velocity is zero $(u=0).$
\[h=ut+\frac{1}{2}gt^2\] \[h=0+\frac{1}{2}gt^2\] \[\therefore t=\sqrt{\frac{2h}{g}}\]
If the new height is doubled the original, $h’=2h.$
Let $t’$ be the new required time. \[t’=\sqrt{\frac{2h’}{g}}=\sqrt{\frac{2(2h)}{g}}=\sqrt{2}\sqrt{\frac{2h}{g}}=\sqrt{2}\;t\]
Thus, the new time will be $\sqrt{2}$ times the original time.
[Equations used above are deduced from Motion Under Gravity]
